Tag: fundamental and derived units

Questions Related to fundamental and derived units

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the solution is:

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 2.11 g/cc

  2. 1.11 g/cc

  3. 12.2 g/cc

  4. 20.4 g/cc

Show answer
Correct Option: B
Explanation:

${ \rho  } _{ mixture }=\dfrac { { \rho  } _{ 1 }{ V } _{ 1 }+{ \rho  } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $


               $=\dfrac { 1.2\times 40+1\times 30 }{ 70 } =\dfrac { 78 }{ 70 } $


${ \rho  } _{ mixture }=1.11gm/cc$

The densities of three liquids are D, 2D and 3D. What will be the density of the resulting mixture if equal volumes of the three liquids are mixed? 

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 6D

  2. 1.4D

  3. 2D

  4. 3D

Show answer
Correct Option: C
Explanation:

let $V$ be the volume of each liquid then the total volume of the mixture becomes $3V$. 

$\rho = \dfrac{total\ mass}{total\ volume}$

Therefore, the mass of the liquids can be written as:
$m _1=D\times V=DV$
$m _1=2D\times V=2DV$
$m _1=3D\times V=3DV$

the total mass of the liquids is
$M= DV+2DV+3DV=6DV$

Therefore, the density of the mixture is:
$\rho=\dfrac{6DV}{3V}$

$\rho = 2D$

One litre of cool air weighs heavier than one litre of hot air because of the : 

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. Increased number of collisions between . the molecules

  2. Increased number of molecules at. low temperature

  3. Greater energy of molecules at high temperature

  4. Lower energy of molecules at high temperature

Show answer
Correct Option: B
Explanation:

Their exists less no. of molecules at the high temperature and high no. of molecules at the low temperature i.e. in the cold conditions.

$60  cc$ of a liquid of relative density $1.4$ are mixed with $40  cc$ of another liquid of relative density $0.8$. The density of the mixture is

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $1.16 {g}/{cc}$

  2. $2.26 {g}/{cc}$

  3. $11.6 {g}/{cc}$

  4. $116 {g}/{cc}$

Show answer
Correct Option: A
Explanation:
given : - ${ S } _{ 1 }=1.4\quad \quad \quad \quad { V } _{ 1 }=60cc$
               ${ S } _{ 2 }=0.89/cc\quad \quad { V } _{ 2 }=40cc$


The density of the mixture is given by:
$S _{mixture }= \dfrac { { S } _{ 1 }{ V } _{ 1 }+{ S } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $

$S _{mixture} = \dfrac { 1.4\times 60+0.8\times 40 }{ 100 } $

$S _{mixture} = 1.16 g/cc$

If two masses A and B have their masses in the ratio 1 : 4 and their volumes are equal, then their densities have the ratio :

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 1 : 4

  2. 4 : 1

  3. 2 : 1

  4. 3 : 1

Show answer
Correct Option: A
Explanation:
We know that density is defined as:
$\rho=\dfrac{mass}{volume}$

It means that density is directly proportional to the mass.
$\dfrac{d}{d'}=\dfrac{m}{m'}=\dfrac14$

Since, $m:m'=1:4$

$\therefore, d:d'=1:4$

A wire of length 50 cm has a mass of 20 g. If its radius is halved by stretching, its new mass per unit length will be :

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 0.4 g $cm^{-1}$

  2. 0.2 kg $m^{-1}$

  3. 0.1 g $cm^{-1}$

  4. 0.2 g $cm^{-1}$

Show answer
Correct Option: C
Explanation:

Mass, $m=\rho V=\rho (\pi r^2 l)$
where $\rho=$ density of the material of the wire, $r=$ radius of wire and $l=$ length of wire. 

When its radius is halved by stretching, the new mass will be ,
 $m'=\rho \pi (r/2)^2l'$

Thus, $\dfrac{m'}{m}=\dfrac{l'}{4l}$ or $\dfrac{m'}{l'}=\dfrac{m}{4l}$

Hence, the new mass per unit length $=m'/l'=m/4l=\dfrac{20}{4\times 50}=0.1 $ $g$ $cm^{-1}$

Ventilators are provided at the top of room

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. To bring oxygen for breathing

  2. So that sunlight may enter the room

  3. To maintain convectional currents to keep the air fresh in the room

  4. To provide an outlet for carbon dioxide

Show answer
Correct Option: C
Explanation:

Ventilators are provided in the rooms at the top of the roofs because if the air inside the room gets hot, the hot air rises up and flows through these ventilators and thus cool air remains at bottom. It brings cool and fresh air in the room. Thus ventilators maintain conventional currents to keep the air fresh in the room.

A vessel contains a mixture consisting of ${m} _{1}=7kg$ of nitrogen $\left( { M } _{ 1 }=28 \right) $ and ${m} _{2}=11g$ of carbon dioixide $\left( { M } _{ 2 }=44 \right) $ at temeprature $T=300K$ and pressure ${ P } _{ 0 }=1\quad atm$. The density of the mixture is:

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $1.446g$ per litres

  2. $2.567g$ per litre

  3. $3.752g$ per litre

  4. $4.572g$ per litre

Show answer
Correct Option: A
Explanation:

Let V is the volume of the vessel.

Now, let $p _{1}$ and $p _{2}$ be the partial pressure, then using gas law: 

$p _{1}V = \dfrac{m _1}{M _1}RT\\$

$p _{2}V = \dfrac{m _2}{M _2}RT,\ p _{0}  = p _{1} +  p _{2}\\$

$p _{0} = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{V}\\$

$V = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{p _{0}}\\$

$\because \rho _{mix}=\dfrac{(m _{1} + m _{2})}{V}\\$

$rho _{mix}=\dfrac {(m _1 + m _2)M _1 M _2} {(m _1M _2 + m _2M _1)} \times \dfrac{p _0}{RT}\\$

Substituting values,

$\rho _{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$

$= 1.446 \ per \ litre$

Option A is correct.

Two non-mixing liquids of densities $\rho$ and $n \rho ( n >$ 1) are put in a container. The height of each liquids $h$ . A solid cylinder of length $L$ and density $d$ is put in this container. The cylinder floats with its its axis vertical and length $p L ( p < 1 )$ in the denser liquid. The density $d$ is equal to

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. ${ 1 + ( n - 1 ) p } p$

  2. ${ 1 + ( n + 1 ) p } p$

  3. ${ 2 + ( n + 1 ) p } p$

  4. ${ 2 - ( n + 1 ) p } p$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} d=density\, \, of\, \, cylin{ { de } }r \ A=area\, \, of\, \, cross-sectional\, \, of\, \, cylinder \ U\sin  g\, \, law\, \, of\, \, floation, \ weight\, \, of\, \, cylinder=up\, thrust\, \, by\, \, two\, \, liquids \ L\times A\times d\times g \ =n\rho \times \left( { pL\times A } \right) g+\rho \left( { L-pL } \right) Ag \ d=np\rho +\rho \left( { 1-p } \right) =\left( { np+1-p } \right) \rho  \ d=\left{ { 1+\left( { n-1 } \right) p } \right} \rho  \end{array}$

Hence,
option $(A)$ is correct answer.

Calculate the mass of air enclosed in a room of length, breadth and height equal to $5m,3m$ and $4m$ respectively.
Density of air $=1.3kg/{m}^{3}$.
chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 60 kg

  2. 78 kg

  3. 18 kg

  4. 10 kg

Show answer
Correct Option: B
Explanation:

The density of a body is given by:

$density=\dfrac{mass}{volume}$

$1.3=\dfrac{mass}{(5 \times 3 \times 4)}$


$mass=1.3\times 60=78kg$